A driver leaves is city at 6 a.m. and another driver the city from the same place at 11:30 a.m. in the same direction . if the average speed of the first driver was 55 km/hour and the second driver 65 km/hour. find the time at which they will be together.
Answers
Answer:
A driver leaves is city at 6 a.m
before another driver starts from the city first driver travels x kms
s=d/t
55=d/5.5hrs
d=302.5 kms
meeting time of both the drivers= distance/relative speed
time=302.5/7
43.21hrs
answer=11.30 am + 43.21 hours
Step-by-step explanation:
Explanation
Lets assume, they will be together in 't' hours after 6 a.m. when the first driver leaves the city.
As the second driver leaves at 11:30 a.m., that means he leaves (11:30am - 6am)= 5.5 hours after the first driver. So, the time duration for second driver is (t- 5.5) hours.
The average speed of the first driver was 55 km/hour and the second driver 65 km/hour.
As Distance= Speed × Time , so the equation will be ....
\begin{gathered}55t=65(t-5.5)\\ \\ 55t=65t-357.5\\ \\ 55t-65t=-357.5\\ \\ -10t=-357.5\\ \\ t=\frac{-357.5}{-10}=35.75\end{gathered}
55t=65(t−5.5)
55t=65t−357.5
55t−65t=−357.5
−10t=−357.5
t=
−10
−357.5
=35.75
So, they will be together after 35.75 hours or 35 hours 45 minutes from 6 am , which is at (6 am+ 35 hours 45 minutes)= 5:45 p.m. on the next day.